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3 years ago

6

A truck drives 80 km in 2 hours. The truck stops for 1 hr and continues driving for 100

1 answer:

bulgar [2K]3 years ago

7 0

Answer:

It Crashed Because it was speeding LOL

Explanation:

It was speeding so cops put spices and truck Crashed

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Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision

RideAnS [48]

Answer:8 m/s

Explanation:

Given

$m_1=1 kg$

$m_2=7 kg$

kinetic Energy of $m_1=32 J$

initially $m_2$ is at rest and let say $m_1$ is moving with velocity u

kinetic Energy of $m_1 is =\frac{m_1u^2}{2}=32$

$u^2=64$

$u=8 m/s$

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

$m_1u+0=(m_1+m_2)v$

$v=\frac{1\times 8}{1+7}=1 m/s$

therefore Final velocity with which both blocks moves is 1 m/s

7 0

3 years ago

Which is the quickest way to accurately take your heart rate

Thepotemich [5.8K]

I would say by putting two fingers under your chin or putting two fingers on the back of your wrist, hope i helped ! :)

3 0

3 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

3 years ago

a constant force acts on an object of mass 10 kg for a duration of 2 seconds it increases the object velocity from 4metre per se

inn [45]

Answer:

F=ma

a=v/t

a=vf-vi/t

a=8-4/2

a=4÷2

a=2msec*2

F=10×2

F=20N

3 0

2 years ago